Topic: [Brain Teaser] Three doors (Read 2338 times)
Yes you should switch it.
But there could be nothing at all right?
No, there is always a prize behind one of the doors.
And the others have nothing, or 1 satoshi like OP said, or a goat xD
Behind one door is 1BTC, behind the others, 1 satoshi. You pick a door, say #1, but you can't open it yet.
But there could be nothing at all right?
This is the MONTY HALL PROBLEM. Just Google it.
@(oYo) I'm not going to insist more after this, but let me just say this: you can't see the problem as being separate in before and after the host opens the doors. If you do it's like you're starting the game at that time. But you didn't, you have to start when you choose the first door. And you have to see that the host doesn't choose 2 random doors. He chooses your door to leave closed and another one that has the prize if you missed. Your door here has 33% chance of having the prize.
So when the host opens the empty door, you are right in saying that it has 0% chance of having the prize. But you don't distribute its probability equally to the 2 doors remaining. You add it to the other door the host chose. That door has 66% chance of having the prize, not 50%. The one you chose has only 33% as before.
If it isn't like this, then how do you explain the image that was posted above and that others talked about already?
Like this you win in 2 out of 3 situations if you change to the other door. So if the position of the car is always random you have a 66% chance of winning.
@criptix ah sorry, I misunderstood what you were saying.
That graphic is a misrepresentation of your choices (like I've already stated), unless of course you're retarded and insist on choosing the revealed door. Just eliminate the line with the revealed door and you'll automatically understand why there's a 1/2 chance, not 1/3. Fudging the percentages to make it seem like one choice now has a 66% chance is a semantic fallacy. Unless you picked the revealed door to begin with, the only thing you accomplish by switching doors is making more work for yourself.So when the host opens the empty door, you are right in saying that it has 0% chance of having the prize. But you don't distribute its probability equally to the 2 doors remaining. You add it to the other door the host chose. That door has 66% chance of having the prize, not 50%. The one you chose has only 33% as before.
If it isn't like this, then how do you explain the image that was posted above and that others talked about already?
That's very clever thanks for sharing this quiz 
. On the Wikipedia link you've shared this graphic explains it well:
I can actually understand how it works now
.
. On the Wikipedia link you've shared this graphic explains it well:I can actually understand how it works now
.Like this you win in 2 out of 3 situations if you change to the other door. So if the position of the car is always random you have a 66% chance of winning.
@criptix ah sorry, I misunderstood what you were saying.
The following riddle is also a masterpiece of equivocation. The trick to solving this riddle is in simply understanding this fact.
Quote
Three men rent a hotel room. Each pays $10 for a total of $30 spent on the room. The next day the hotel owner tells the three men that they over paid for the room as it only costs $25. The three men tell the owner to give them each a dollar back and he can keep two dollars.
If you do the math, each man paid $9 a piece for the room for a total of $27. The owner kept $2 which brings the total to $29.
The question is where did the other dollar go?
If you do the math, each man paid $9 a piece for the room for a total of $27. The owner kept $2 which brings the total to $29.
The question is where did the other dollar go?
@(oYo) I'm not going to insist more after this, but let me just say this: you can't see the problem as being separate in before and after the host opens the doors. If you do it's like you're starting the game at that time. But you didn't, you have to start when you choose the first door. And you have to see that the host doesn't choose 2 random doors. He chooses your door to leave closed and another one that has the prize if you missed. Your door here has 33% chance of having the prize.
So when the host opens the empty door, you are right in saying that it has 0% chance of having the prize. But you don't distribute its probability equally to the 2 doors remaining. You add it to the other door the host chose. That door has 66% chance of having the prize, not 50%. The one you chose has only 33% as before.
If it isn't like this, then how do you explain the image that was posted above and that others talked about already?
Like this you win in 2 out of 3 situations if you change to the other door. So if the position of the car is always random you have a 66% chance of winning.
@criptix ah sorry, I misunderstood what you were saying.
So when the host opens the empty door, you are right in saying that it has 0% chance of having the prize. But you don't distribute its probability equally to the 2 doors remaining. You add it to the other door the host chose. That door has 66% chance of having the prize, not 50%. The one you chose has only 33% as before.
If it isn't like this, then how do you explain the image that was posted above and that others talked about already?
That's very clever thanks for sharing this quiz . On the Wikipedia link you've shared this graphic explains it well:
I can actually understand how it works now.
. On the Wikipedia link you've shared this graphic explains it well:I can actually understand how it works now
.Like this you win in 2 out of 3 situations if you change to the other door. So if the position of the car is always random you have a 66% chance of winning.
@criptix ah sorry, I misunderstood what you were saying.
That's very clever thanks for sharing this quiz . On the Wikipedia link you've shared this graphic explains it well:
I can actually understand how it works now.
. On the Wikipedia link you've shared this graphic explains it well:I can actually understand how it works now
. 
Thanks for your contributions!
While most of you already knew about the Monty Hall problem, it's always interesting to discuss, especially how to turn it into something more intuitive. The meta problem of the difference between 'No' and 'It doesn't matter' was an interesting development too
While most of you already knew about the Monty Hall problem, it's always interesting to discuss, especially how to turn it into something more intuitive. The meta problem of the difference between 'No' and 'It doesn't matter' was an interesting development too
There are two separate instances (equations) here, of which you are trying to represent as one.
The first round consisted of a 1/3 chance, on account there were 3 doors. That round ended with no winners, but one of the doors was shown to be a loser.
The next round essentially only had two doors to choose from, ergo there was a 1/2 chance to choose the right door.
Stop trying to insist that the losing door in the first round is even a legitimate consideration anymore. There is a 0% percent chance that that door (or another 997 just like it) is the right choice now, as it has been proven to be the wrong one. It is no longer an option, now. In this new round (equation), consideration is only regarded for two brand new choices essentially. Implying there are any other choices or odds is simply misleading.
So, the correct answer to the poll's question should be "No, it doesn't matter." "No" there is no advantage to switching, nor is there any disadvantage, as "it doesn't matter" which door you pick, since they both have a 50% chance of winning, now.
TL;DR - Ok, here's a great graphic to help make my point. Pick the "not a goat" door. Here's 3 doors to choose from. There's a goat behind 2 of them. Door number 3 absolutely has a goat and one of the other 2 doesn't. Picking door 3 is an automatic loss, so if you want a (50%) chance to win, you will only pick door 1 or door 2.
The first round consisted of a 1/3 chance, on account there were 3 doors. That round ended with no winners, but one of the doors was shown to be a loser.
The next round essentially only had two doors to choose from, ergo there was a 1/2 chance to choose the right door.
Stop trying to insist that the losing door in the first round is even a legitimate consideration anymore. There is a 0% percent chance that that door (or another 997 just like it) is the right choice now, as it has been proven to be the wrong one. It is no longer an option, now. In this new round (equation), consideration is only regarded for two brand new choices essentially. Implying there are any other choices or odds is simply misleading.
So, the correct answer to the poll's question should be "No, it doesn't matter." "No" there is no advantage to switching, nor is there any disadvantage, as "it doesn't matter" which door you pick, since they both have a 50% chance of winning, now.
TL;DR - Ok, here's a great graphic to help make my point. Pick the "not a goat" door. Here's 3 doors to choose from. There's a goat behind 2 of them. Door number 3 absolutely has a goat and one of the other 2 doesn't. Picking door 3 is an automatic loss, so if you want a (50%) chance to win, you will only pick door 1 or door 2.
I heard the solution in the film "Las Vegas 21"
^ I thought with more doors would be more intuitive. : /
With 1000 doors, when you choose a door in the beginning you only have 0.1% probability of choosing the right one. The game host knows where the prize is and opens every door except the one you choose and another one.
But your choice was still made with a 0.1% chance. It doesn't change to 50% now, because that would be losing information. You could only say it was 50% chance if you started from 2 doors, not 1000.
With 1000 doors, when you choose a door in the beginning you only have 0.1% probability of choosing the right one. The game host knows where the prize is and opens every door except the one you choose and another one.
But your choice was still made with a 0.1% chance. It doesn't change to 50% now, because that would be losing information. You could only say it was 50% chance if you started from 2 doors, not 1000.
yes, that was exactly what i was trying to point out.
people think of it as there would be only 2 doors to choose from and then logically the probality would be 50%
^ I thought with more doors would be more intuitive. : /
With 1000 doors, when you choose a door in the beginning you only have 0.1% probability of choosing the right one. The game host knows where the prize is and opens every door except the one you choose and another one.
But your choice was still made with a 0.1% chance. It doesn't change to 50% now, because that would be losing information. You could only say it was 50% chance if you started from 2 doors, not 1000.
With 1000 doors, when you choose a door in the beginning you only have 0.1% probability of choosing the right one. The game host knows where the prize is and opens every door except the one you choose and another one.
But your choice was still made with a 0.1% chance. It doesn't change to 50% now, because that would be losing information. You could only say it was 50% chance if you started from 2 doors, not 1000.
I would agree that oYo. I have to agree with this analysis. Once you know that one of the other doors is worthless you have a 50% chance of having picked the right door. Just like if or when you find out you lost, you can say you have a 0% chance of having picked the right door, regardless of how good your chances were when you didn't have that knowledge.
In other words, it doesn't matter whether you switch or not.
sdp
In other words, it doesn't matter whether you switch or not.
sdp
Lol. Maybe this is better to help see why that is wrong: imagine you have 1000 doors instead of 3.
You pick 1 out of 1000, and the host opens 998 doors that don't have the prize. Would you change to the door he left then or keep the same? xD
I think right now its not about mathematical probalities but about the game show.
if the moderator opens 1 out of the three (998 out of 1000) and we know for sure it is not the prize then only two doors remain.
1 out of 2 is 50%
Like you would not need to pick at all in the first place
I would agree that oYo. I have to agree with this analysis. Once you know that one of the other doors is worthless you have a 50% chance of having picked the right door. Just like if or when you find out you lost, you can say you have a 0% chance of having picked the right door, regardless of how good your chances were when you didn't have that knowledge.
In other words, it doesn't matter whether you switch or not.
sdp
In other words, it doesn't matter whether you switch or not.
sdp
Lol. Maybe this is better to help see why that is wrong: imagine you have 1000 doors instead of 3.
You pick 1 out of 1000, and the host opens 998 doors that don't have the prize. Would you change to the door he left then or keep the same? xD
I would agree that oYo. I have to agree with this analysis. Once you know that one of the other doors is worthless you have a 50% chance of having picked the right door. Just like if or when you find out you lost, you can say you have a 0% chance of having picked the right door, regardless of how good your chances were when you didn't have that knowledge.
In other words, it doesn't matter whether you switch or not.
sdp
In other words, it doesn't matter whether you switch or not.
sdp
I read the explanation on wikipedia a while back. Still doesn't feel right, and I will never have an intuitive understanding of it I suppose.
I think of it like dice at times , and its like saying that select a range between 1 and 100, between 1-33,34-67,68-100(dont worry about the differnce of 1, just consider it 1/3) . Now you select 1-33. Dice site owner checks seed to tell you its not 68-100. Then how does switching to 34-67 increase odds ?
I think of it like dice at times , and its like saying that select a range between 1 and 100, between 1-33,34-67,68-100(dont worry about the differnce of 1, just consider it 1/3) . Now you select 1-33. Dice site owner checks seed to tell you its not 68-100. Then how does switching to 34-67 increase odds ?
There is an example to make it easier: You have 3 doors, Red Yellow and Green.
In the case that the Bitcoin us behind the red door:
If you pick the red door switching would make you lose
If you pick the yellow door switching would make you win
If you pick the green door switching would make you win
Lose 1/3
Win 2/3
Thanks that does clear it up. But with the knowledge of this, the game show host might trick you into a trap , and you may actually lose, if you had selected the red from the start.
Fuck btc
I want my goat
I want my goat
My initial choice has a 33% chance of being right. (You could also say there's a 66% chance of choosing the wrong door.) That's a gimme, since there are initially 3 doors to choose from.
If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one. Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.
This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.
If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one. Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.
This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.
You should switch to the other door. xD
I don't want to spoil but you can't ignore the 33% chance you had in the beginning. That didn't change to 50%. Those are different things.
OyO is wrong, the chances of winning are 66% if you switch, it has been explained already with this example:
If the red door is the good door you have 3 possibilities:
1 pick red door and lose if you switch
2 pick yellow door and win if you switch
3 pick green door and win if you switch
Win 2/3 and lose 1/3 so 66%
Yes and that explanation was posted before his answer and he didn't read it or didn't care lol. So I was using his answer to explain where he went wrong.
By the way, is Bardman your alt account then? xD
My initial choice has a 33% chance of being right. (You could also say there's a 66% chance of choosing the wrong door.) That's a gimme, since there are initially 3 doors to choose from.
If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one. Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.
This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.
If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one. Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.
This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.
You should switch to the other door. xD
I don't want to spoil but you can't ignore the 33% chance you had in the beginning. That didn't change to 50%. Those are different things.
OyO is wrong, the chances of winning are 66% if you switch, it has been explained already with this example:
If the red door is the good door you have 3 possibilities:
1 pick red door and lose if you switch
2 pick yellow door and win if you switch
3 pick green door and win if you switch
Win 2/3 and lose 1/3 so 66%
My initial choice has a 33% chance of being right. (You could also say there's a 66% chance of choosing the wrong door.) That's a gimme, since there are initially 3 doors to choose from.
If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one. Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.
This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.
If one of the doors (which I didn't choose) is revealed to be a poor choice and subsequently removed from the equation, that leaves 2 doors to choose from, ergo there's a 50% chance now of choosing the right (or wrong) one. Since both of the remaining doors now have an equal chance of being the right (or better) choice, this also means that my odds of having already chosen the right door just shot up from 33% to 50%. Another way of looking at it is that my odds of choosing the wrong door went down from 66% to 50%. Hence there's absolutely no advantage (or disadvantage) to switching to the other door.
This 'brain teaser' feels somewhat like that joke where the three guys get a rebate on their hotel room and you are misdirected to account for a missing dollar.
You should switch to the other door. xD
I don't want to spoil but you can't ignore the 33% chance you had in the beginning. That didn't change to 50%. Those are different things.
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